Question

A vertical spring (spring constant =160 N/m) is mounted on the floor. A 0.340-kg block is placed on top of the spring and pushed down to start it oscillating in simple harmonic motion. The block is not attached to the spring. (a) Obtain the frequency (in Hz) of the motion. (b) Determine the amplitude at which the block will lose contact with the spring.

Answer 1

(a) 3.5 Hz

The angular frequency in a spring-mass system is given by

$\omega=\sqrt{\frac{k}{m}}$

where

k is the spring constant

m is the mass

Here in this problem we have

k = 160 N/m

m = 0.340 kg

So the angular frequency is

$\omega=\sqrt{\frac{160 N/m}{0.340 kg}}=21.7 rad/s$

And the frequency of the motion instead is given by:

$f=\frac{\omega}{2\pi}=\frac{21.7 rad/s}{2\pi}=3.5 Hz$

(b) 0.021 m

The block is oscillating up and down together with the upper end of the spring. The block will lose contact with the spring when the direction of motion of the spring changes: this occurs when the spring is at maximum displacement, so at

x = A

where A is the amplitude of the motion.

The maximum displacement is given by Hook's law:

$F=kA$

where

F is the force applied initially to the spring, so it is equal to the weight of the block:

$F=mg=(0.340 kg)(9.81 m/s^2)=3.34 N$

k = 160 N/m is the spring constant

Solving for A, we find

$A=\frac{F}{k}=\frac{3.34 N}{160 N/m}=0.021 m$